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The following mechanism has been proposed for the gasphase reaction of chloroform (CHCl \(_{3}\) ) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \stackrel{\mathrm{k}_{1}}{\mathrm{k}_{-1}} 2 \mathrm{Cl}(g)\) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g)\) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{CCl}_{4}\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism?

Short Answer

Expert verified
(a) Overall: \(\mathrm{CHCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(g) + \mathrm{HCl}(g)\). (b) Intermediates: \(\mathrm{Cl}\) and \(\mathrm{CCl}_{3}\). (c) Molecularity: Step 1, 2, and 3 are bimolecular. (d) Rate-determining: Step 2. (e) Rate law: \(\text{Rate} = k [\mathrm{Cl}_{2}]^{1/2}[\mathrm{CHCl}_{3}]\).

Step by step solution

01

Identify the Overall Reaction

To find the overall reaction, add all the elementary steps given in the mechanism. Cancel out any species that appear on both sides of the reaction equations. Here, \(2 \mathrm{Cl}(g)\), \( \mathrm{CCl}_{3}(g)\), and \( \mathrm{HCl}(g)\) cancel out from reactant and product sides respectively. The net reaction results in: \[\mathrm{CHCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(g) + \mathrm{HCl}(g).\]
02

Identify the Intermediates

Intermediates are species that are formed and consumed during the course of the mechanism but do not appear in the overall reaction. From the steps, \(\mathrm{Cl}(g)\) and \(\mathrm{CCl}_{3}(g)\) qualify as intermediates since they are formed and used up within the reaction mechanism.
03

Determine Molecularity of Each Step

Molecularity refers to the number of reactant molecules involved in an elementary reaction. Step 1 is bimolecular as it involves \(\mathrm{Cl}_{2}\) dissociating into two \(\mathrm{Cl}\) atoms. Step 2 is also bimolecular as it involves \(\mathrm{Cl}(g)\) reacting with \(\mathrm{CHCl}_{3}(g)\). Step 3 is bimolecular as \(\mathrm{Cl}(g)\) reacts with \(\mathrm{CCl}_{3}(g)\).
04

Identify the Rate-Determining Step

The rate-determining step is the slowest step in the reaction mechanism and limits the rate of the overall reaction. Here, Step 2 involving \(\mathrm{Cl}(g)\) and \(\mathrm{CHCl}_{3}(g)\) is the slow step, thus it is the rate-determining step.
05

Predict the Rate Law from Mechanism

The rate law for a reaction is derived from the rate-determining step. Since Step 2 is the rate-determining step, its rate law can be expressed as \(\text{Rate} = k_2 [\mathrm{Cl}][\mathrm{CHCl}_{3}]\). To express rate entirely in terms of initial reactants, substitute \([\mathrm{Cl}]=\frac{k_1}{k_{-1}} [\mathrm{Cl}_{2}]\) from Step 1, which gives: \[\text{Rate} = k [\mathrm{Cl}_{2}]^{1/2}[\mathrm{CHCl}_{3}].\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Overall Reaction
In chemical reactions, especially those involving multiple steps, identifying the overall reaction is crucial. The overall reaction encompasses all substances consumed and produced but leaves out any intermediates. This means we look at the starting compounds and the end products. In the given sequence of reactions between chloroform (CHCl3) and chlorine (Cl2), the overall reaction is determined by adding up all the individual steps. We start with CHCl3 and Cl2 and after all the intermediate steps, we end with CCl4 and HCl. Thus, the overall reaction is:\[\mathrm{CHCl}_{3}(g) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(g) + \mathrm{HCl}(g).\]This reaction shows the conversion of chloroform and chlorine into carbon tetrachloride and hydrochloric acid.
Intermediates
Intermediates play an essential role in multi-step reaction mechanisms. They are species that are generated in one step and consumed in another. Thus, intermediates do not appear in the overall chemical equation. They "bridge the gap" between reactants and products without being present in the final equation. In the proposed mechanism, the intermediates are Cl(g) and CCl3(g). These intermediates are formed during the breakdown of Cl2 and are used up in subsequent steps. They facilitate the reaction but do not impact the overall stoichiometry of the starting and ending compounds.
Molecularity
Molecularity is a concept that describes the number of molecules participating in an elementary reaction step. It is different from order of a reaction, as molecularity refers only to elementary reactions and is always an integer. In the provided mechanism, each step has different molecularity:
  • Step 1 has a molecularity of two (bimolecular) because it involves the dissociation of Cl2 into two Cl atoms.
  • Step 2 also has a molecularity of two, involving Cl reacting with CHCl3.
  • Step 3 is again bimolecular as it involves Cl reacting with CCl3 to form the final product CCl4.
Understanding molecularity helps in predicting how reaction rates might change under different concentrations of reactants in elementary steps.
Rate-Determining Step
The rate-determining step (RDS) is the slowest step in a reaction mechanism. It acts as a bottleneck, controlling the overall speed of the reaction. It's like the narrowest part of a funnel slowing down the whole process. Identifying this step is crucial because the kinetics of the entire reaction depend on it. In the chloroform-chlorine reaction mechanism, Step 2 is identified as the rate-determining step. This step involves the reaction between Cl(g) and CHCl3(g), highlighting that the formation of CCl3 and HCl is the slowest, and thus limits the speed at which products form. Knowing the RDS allows chemists to devise strategies to increase the overall reaction rate, such as using catalysts.
Rate Law
The rate law describes how the rate of a chemical reaction depends on the concentration of reactants. For complex reactions, the rate law is derived from the rate-determining step. In our scenario, since Step 2 is the slowest step, its equation gives the rate law:\[ \text{Rate} = k_2 [\mathrm{Cl}][\mathrm{CHCl}_{3}] \]However, since Cl is an intermediate, we express its concentration in terms of observable reactants. By considering the equilibrium in Step 1, we can substitute \[ [\mathrm{Cl}] = \frac{k_1}{k_{-1}} [\mathrm{Cl}_{2}]^{1/2} \]into the rate law to adjust for intermediates concentration:\[ \text{Rate} = k [\mathrm{Cl}_{2}]^{1/2}[\mathrm{CHCl}_{3}] \]This equation predicts how changes in the concentrations of Cl2 and CHCl3 affect the reaction rate, thus guiding the optimization of reaction conditions.

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Most popular questions from this chapter

(a) Develop an equation for the half-life of a zero-order reaction. (b) Does the half-life of a zero-order reaction increase, decrease, or remain the same as the reaction proceeds?

From the following data for the second-order gas-phase decomposition of HI at \(430^{\circ} \mathrm{C},\) calculate the second-order rate constant and half- life for the reaction: \begin{tabular}{rl} \hline Time (s) & [HIYmol dm \(^{-3}\) \\ \hline 0 & 1 \\ 100 & 0.89 \\ 200 & 0.8 \\ 300 & 0.72 \\ 400 & 0.66 \end{tabular}

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The rate of disappearance of HCl was measured for the following reaction: $$ \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ The following data were collected: \begin{tabular}{cc} \hline Time (min) & {\([\mathrm{HCl}](M)\)} \\ \hline 0.0 & 1.85 \\ 54.0 & 1.58 \\ 107.0 & 1.36 \\ 215.0 & 1.02 \\ 430.0 & 0.580 \\ \hline \end{tabular} (a) Calculate the average rate of reaction, in \(M / \mathrm{s}\), for the time interval between each measurement. (b) Calculate the average rate of reaction for the entire time for the data from \(t=0.0 \mathrm{~min}\) to \(t=430.0 \mathrm{~min} .(\mathbf{c})\) Which is greater, the average rate between \(t=54.0\) and \(t=215.0 \mathrm{~min}\), or between \(t=107.0\) and \(t=430.0 \mathrm{~min} ?\) (d) Graph \([\mathrm{HCl}\) versus time and determine the instantaneous rates in \(M / \min\) and \(M /\) s at \(t=75.0 \mathrm{~min}\) and \(t=250 \mathrm{~min} .\)

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